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AlgebraQuestion and Answers: Page 299

Question Number 70460 Answers: 0 Comments: 2

Question Number 70312 Answers: 1 Comments: 1

If, (1/a^2 )+(1/b^2 )+(1/c^2 ) = (1/(ab))+(1/(bc))+(1/(ca)) then prove that, a=b=c.

$$\mathrm{If},\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{ab}}+\frac{\mathrm{1}}{\mathrm{bc}}+\frac{\mathrm{1}}{\mathrm{ca}}\:\mathrm{then}\:\mathrm{prove}\: \\ $$$$\mathrm{that},\:\mathrm{a}=\mathrm{b}=\mathrm{c}. \\ $$

Question Number 70216 Answers: 0 Comments: 0

let A = (((1 −1)),((1 2)) ) 1)calculate A^n 2) find e^A and e^(−A)

$${let}\:{A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\end{pmatrix} \\ $$$$\left.\mathrm{1}\right){calculate}\:{A}^{{n}} \:\:\:\:\: \\ $$$$\left.\mathrm{2}\right)\:{find}\:{e}^{{A}} \:{and}\:{e}^{−{A}} \\ $$

Question Number 70162 Answers: 1 Comments: 0

Question Number 70161 Answers: 1 Comments: 0

Question Number 70135 Answers: 0 Comments: 1

sophie−Germain identity a^4 +4b^4 =((a+b)^2 +b^2 )((a−b)^2 +b^2 )

$${sophie}−{Germain}\:{identity} \\ $$$${a}^{\mathrm{4}} +\mathrm{4}{b}^{\mathrm{4}} =\left(\left({a}+{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\left({a}−{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \right) \\ $$

Question Number 70103 Answers: 2 Comments: 0

if m^3 +2p^3 =3mn, a^3 +b^3 =p^3 and a^2 +b^2 =n then prove that a+b=m.

$$\mathrm{if}\:\mathrm{m}^{\mathrm{3}} +\mathrm{2p}^{\mathrm{3}} =\mathrm{3mn},\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} =\mathrm{p}^{\mathrm{3}} \:\mathrm{and} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{n}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{a}+\mathrm{b}=\mathrm{m}. \\ $$

Question Number 70040 Answers: 1 Comments: 3

If, a^2 b^2 c^2 ((1/a^3 )+(1/b^3 )+(1/c^3 ))=a^3 +b^3 +c^3 than prove that a,b,c Successive Proportional.

$$\mathrm{If},\:\mathrm{a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{3}} }\right)=\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} \:\mathrm{than} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{Successive}\:\mathrm{Proportional}. \\ $$

Question Number 69995 Answers: 4 Comments: 0

1. { (((√(x^2 +y^2 ))+(√(x^2 −y^2 ))=a)),(((√(x+y))+(√(x−y))=b)) :} [a,b∈R^+ ] 2. { (((√((√x)+y))+(√(x+(√y)))=a)),(((√((√x)−y))+(√(x−(√y)))=b)) :} 3. { ((x^2 +y^2 =(a−b)xy)),((x^3 +y^3 =ab(x−y))) :} 4. { ((x+y+z=a)),((x^2 +y^2 +z^2 =b)),((x^3 +y^3 +z^3 =abxyz)) :} 5. (√(x^2 +(√x)))+(√(x^2 −(√x)))=2 6. (√(x^2 +x+1))+(√(x^2 +x−1))+(√(x^2 −x−1))=1 7. 16^(sin^2 x) +16^(cos^2 x) =10 8. 2^(lnx) =x.ln(x+(√2))

Question Number 69939 Answers: 2 Comments: 12

x,y,z ∈ Z^+ find all solutions of xy=(x+y)z

$${x},{y},{z}\:\in\:{Z}^{+} \\ $$$${find}\:{all}\:{solutions}\:{of}\:\boldsymbol{{xy}}=\left(\boldsymbol{{x}}+\boldsymbol{{y}}\right)\boldsymbol{{z}} \\ $$

Question Number 69741 Answers: 2 Comments: 4

Question Number 69738 Answers: 1 Comments: 0

Sarah dances everyday of the week, including saturdays and sundays. In november 2018, Sarah had to miss a few days. To control her absences she marks the day she missed class with a x on the calendar. She marked the 5th, 21st and 27th of november. What percentage indicates Sarah′s absences in november?

$${Sarah}\:{dances}\:{everyday}\:{of}\:{the}\:{week}, \\ $$$${including}\:{saturdays}\:{and}\:{sundays}. \\ $$$${In}\:{november}\:\mathrm{2018},\:{Sarah}\:{had}\:{to}\:{miss} \\ $$$${a}\:{few}\:{days}.\:{To}\:{control}\:{her}\:{absences} \\ $$$${she}\:{marks}\:{the}\:{day}\:{she}\:{missed}\:{class} \\ $$$${with}\:{a}\:\boldsymbol{{x}}\:{on}\:{the}\:{calendar}. \\ $$$${She}\:{marked}\:{the}\:\mathrm{5}{th},\:\mathrm{21}{st}\:{and}\:\mathrm{27}{th} \\ $$$${of}\:{november}. \\ $$$${What}\:{percentage}\:{indicates}\:{Sarah}'{s} \\ $$$${absences}\:{in}\:{november}? \\ $$

Question Number 71696 Answers: 1 Comments: 0

The side of a square is measured to be 12cm long cofrect to the nearest cm. Find the maximum absolute error and the maximum percentage error for (a) The length of the square (Answer: 0.5cm, 4.17%) (b) The area of the square. (Answer: 12.25cm, 8.5%)

$$\mathrm{The}\:\mathrm{side}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}\:\mathrm{is}\:\mathrm{measured}\:\mathrm{to}\:\mathrm{be}\:\:\mathrm{12cm}\:\mathrm{long}\:\mathrm{cofrect} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{nearest}\:\:\mathrm{cm}.\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{absolute}\:\mathrm{error} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{percentage}\:\mathrm{error}\:\mathrm{for} \\ $$$$\left(\mathrm{a}\right)\:\:\:\:\mathrm{The}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\:\:\:\:\left(\mathrm{Answer}:\:\:\mathrm{0}.\mathrm{5cm},\:\:\mathrm{4}.\mathrm{17\%}\right) \\ $$$$\left(\mathrm{b}\right)\:\:\:\:\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}.\:\:\:\:\:\:\left(\mathrm{Answer}:\:\:\:\:\mathrm{12}.\mathrm{25cm},\:\:\:\mathrm{8}.\mathrm{5\%}\right) \\ $$

Question Number 69681 Answers: 2 Comments: 2

Question Number 69665 Answers: 1 Comments: 1

Question Number 69662 Answers: 1 Comments: 0

prove that ((2x^3 −x^2 −2x+1)/(x^3 +1)) + ((x^3 +1)/(x^4 −2x^3 +3x^2 −2x+1)) = 2

$${prove}\:{that} \\ $$$$ \\ $$$$\frac{\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{3}} +\mathrm{1}}\:+\:\frac{{x}^{\mathrm{3}} +\mathrm{1}}{{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}\:=\:\mathrm{2} \\ $$

Question Number 69645 Answers: 1 Comments: 0

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Question Number 69538 Answers: 0 Comments: 0

Hello Verry Nice Day for You Find Σ_(k≥0) (1/((8k+1)^2 ))

$${Hello}\:{Verry}\:{Nice}\:{Day}\:{for}\:\:{You} \\ $$$${Find}\:\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{8}{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Question Number 69586 Answers: 0 Comments: 0

x^5 −x^4 −x^3 −x^2 −x−1=0

$${x}^{\mathrm{5}} −{x}^{\mathrm{4}} −{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$

Question Number 69496 Answers: 0 Comments: 5

Question Number 69589 Answers: 0 Comments: 0

x^3 +px−ry+qz+a=0 y^3 +rx+qy−pz+b=0 z^3 −qx+py+rz+c=0 solve for x,y,z, in terms of p,q,r, a,b,c.

$${x}^{\mathrm{3}} +{px}−{ry}+{qz}+{a}=\mathrm{0} \\ $$$${y}^{\mathrm{3}} +{rx}+{qy}−{pz}+{b}=\mathrm{0} \\ $$$${z}^{\mathrm{3}} −{qx}+{py}+{rz}+{c}=\mathrm{0} \\ $$$${solve}\:{for}\:{x},{y},{z},\:{in}\:{terms}\:{of} \\ $$$${p},{q},{r},\:{a},{b},{c}. \\ $$

Question Number 69479 Answers: 0 Comments: 2

to Sir Aifour: we can construct polynomes of both 3^(rd) and 4^(th) degree in a way that the constants are ∈Z or ∈Q and the solutions are not trivial i.e. (t−α)(t+(α/2)−(√β))(t+(α/2)+(√β))=0∧t=x+(γ/3) ⇔ x^3 +γx^2 −(((3α^2 )/4)+β−(γ^2 /3))x−((α^3 /4)+((α^2 γ)/4)−αβ+((βγ)/3)−(γ^3 /(27)))=0 or the more complicated with sinus/cosinus (x−α−(√β)−(√γ)−(√δ))(x−α−(√β)+(√γ)+(√δ))(x−α+(√β)−(√γ)+(√δ))(x−α+(√β)+(√γ)−(√δ))=0 where all constants ∈Q if (√(βγδ))∈Q I could not find a similar construction for a polynome of 5^(th) degree, where the 5 roots are of comparable complexity [(x−a)(x−b−ci)(x−b+ci)(x−d−ei)(x−d+ei) doesn′t count] maybe you should at first focus on this

$$\mathrm{to}\:\mathrm{Sir}\:\mathrm{Aifour}: \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{construct}\:\mathrm{polynomes}\:\mathrm{of}\:\mathrm{both}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{and} \\ $$$$\mathrm{4}^{\mathrm{th}} \:\mathrm{degree}\:\mathrm{in}\:\mathrm{a}\:\mathrm{way}\:\mathrm{that}\:\mathrm{the}\:\mathrm{constants}\:\mathrm{are} \\ $$$$\in\mathbb{Z}\:\mathrm{or}\:\in\mathbb{Q}\:\mathrm{and}\:\mathrm{the}\:\mathrm{solutions}\:\mathrm{are}\:\mathrm{not}\:\mathrm{trivial} \\ $$$$\mathrm{i}.\mathrm{e}. \\ $$$$\left({t}−\alpha\right)\left({t}+\frac{\alpha}{\mathrm{2}}−\sqrt{\beta}\right)\left({t}+\frac{\alpha}{\mathrm{2}}+\sqrt{\beta}\right)=\mathrm{0}\wedge{t}={x}+\frac{\gamma}{\mathrm{3}} \\ $$$$\Leftrightarrow \\ $$$${x}^{\mathrm{3}} +\gamma{x}^{\mathrm{2}} −\left(\frac{\mathrm{3}\alpha^{\mathrm{2}} }{\mathrm{4}}+\beta−\frac{\gamma^{\mathrm{2}} }{\mathrm{3}}\right){x}−\left(\frac{\alpha^{\mathrm{3}} }{\mathrm{4}}+\frac{\alpha^{\mathrm{2}} \gamma}{\mathrm{4}}−\alpha\beta+\frac{\beta\gamma}{\mathrm{3}}−\frac{\gamma^{\mathrm{3}} }{\mathrm{27}}\right)=\mathrm{0} \\ $$$$\mathrm{or}\:\mathrm{the}\:\mathrm{more}\:\mathrm{complicated}\:\mathrm{with}\:\mathrm{sinus}/\mathrm{cosinus} \\ $$$$ \\ $$$$\left({x}−\alpha−\sqrt{\beta}−\sqrt{\gamma}−\sqrt{\delta}\right)\left({x}−\alpha−\sqrt{\beta}+\sqrt{\gamma}+\sqrt{\delta}\right)\left({x}−\alpha+\sqrt{\beta}−\sqrt{\gamma}+\sqrt{\delta}\right)\left({x}−\alpha+\sqrt{\beta}+\sqrt{\gamma}−\sqrt{\delta}\right)=\mathrm{0} \\ $$$$\mathrm{where}\:\mathrm{all}\:\mathrm{constants}\:\in\mathbb{Q}\:\mathrm{if}\:\sqrt{\beta\gamma\delta}\in\mathbb{Q} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{could}\:\mathrm{not}\:\mathrm{find}\:\mathrm{a}\:\mathrm{similar}\:\mathrm{construction}\:\mathrm{for} \\ $$$$\mathrm{a}\:\mathrm{polynome}\:\mathrm{of}\:\mathrm{5}^{\mathrm{th}} \:\mathrm{degree},\:\mathrm{where}\:\mathrm{the}\:\mathrm{5}\:\mathrm{roots} \\ $$$$\mathrm{are}\:\mathrm{of}\:\mathrm{comparable}\:\mathrm{complexity} \\ $$$$\left[\left({x}−{a}\right)\left({x}−{b}−{c}\mathrm{i}\right)\left({x}−{b}+{c}\mathrm{i}\right)\left({x}−{d}−{e}\mathrm{i}\right)\left({x}−{d}+{e}\mathrm{i}\right)\right. \\ $$$$\left.\mathrm{doesn}'\mathrm{t}\:\mathrm{count}\right] \\ $$$$\mathrm{maybe}\:\mathrm{you}\:\mathrm{should}\:\mathrm{at}\:\mathrm{first}\:\mathrm{focus}\:\mathrm{on}\:\mathrm{this} \\ $$

Question Number 69423 Answers: 0 Comments: 1

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